Question

[Linh Nguyen]

1)

\(A=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

a) rút gọn A

b) tìm x nguyên để A nguyên

c) tìm x để \(\left|A\right|=A\)

2)

\(B=\left(\dfrac{3X^3+3}{X^3-1}-\dfrac{X-1}{X^2+X+1}-\dfrac{1}{X-1}\right)\times\dfrac{X-1}{2X^2-5X+5}\)

a) rút gọn B

b) tính GTLN của B

GIÚP MK VS - CẢM ƠN NHIỀU

Answer 1

[Aki Tsuki]

1/ đkxđ: x≠\(\pm\)1; x≠1/2

a/\(A=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

\(=\left(\dfrac{x+1}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(=\dfrac{x+1+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(=\dfrac{2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}=\dfrac{2}{1-2x}\)

b/ A nguyên <=> 1 - 2x ∈ Ư(2)

<=> 1 - 2x = {-2;-1;1;2}

<=> -2x = {-3; -2; 0;1}

<=> x = {3/2; 1; 0; -1/2}

mà x nguyên => x = {1;0}

c/ \(\left|A\right|=A\Leftrightarrow\left|\dfrac{2}{1-2x}\right|=\dfrac{2}{1-2x}\)

+) Với x > 1/2 có:

\(\dfrac{2}{1-2x}=\dfrac{2}{1-2x}\Leftrightarrow\dfrac{2}{1-2x}-\dfrac{2}{1-2x}=0\Leftrightarrow0x=0\)

=> x>1/2 thỏa mãn là nghiệm

+) Với x < 1/2 có:

\(\dfrac{2}{1-2x}=\dfrac{2}{2x-1}\)

\(\Leftrightarrow\dfrac{2}{1-2x}-\dfrac{2}{2x-1}=0\Leftrightarrow\dfrac{2}{1-2x}+\dfrac{2}{1-2x}=0\)

\(\Leftrightarrow\dfrac{4}{1-2x}=0\) mà 1 - 2x ≠ 0 => vô nghiệm

Vậy x>1/2